# How do you find the vertical, horizontal or slant asymptotes for y=(2x^2-3x+4)/(x+2)?

Jun 3, 2016

Vertical asymptotes at ${x}_{v} = - 2$
Slant asymptote $y = 2 x - 7$

#### Explanation:

In a polynomial fraction $f \left(x\right) = \frac{{p}_{n} \left(x\right)}{{p}_{m} \left(x\right)}$ we have:

1) vertical asymptotes for ${x}_{v}$ such that ${p}_{m} \left({x}_{v}\right) = 0$
2) horizontal asymptotes when $n \le m$
3) slant asymptotes when $n = m + 1$
In the present case we have ${x}_{v} = - 2$ and $n = m + 1$ with $n = 2$ and $m = 1$

Slant asymptotes are obtained considering (p_n(x))/(p_{n-1}(x)) approx y = a+b x for large values of $\left\mid x \right\mid$

In the present case we have

$\frac{{p}_{n} \left(x\right)}{{p}_{n - 1} \left(x\right)} = \frac{2 {x}^{2} - 3 x + 4}{x + 2}$
${p}_{n} \left(x\right) = {p}_{n - 1} \left(x\right) \left(a x + b\right) + {r}_{n - 2} \left(x\right)$
$\frac{{p}_{n} \left(x\right)}{{p}_{n - 1} \left(x\right)} = \left(2 {x}^{2} - 3 x + 4\right) = \left(x + 2\right) \left(a x + b\right) + c$

equating we have

{ (4 - 2 b - c=0), (-3 - 2 a - b=0), (2 - a=0) :}

Solving for $a , b , c$ we have $a = 2 , b = - 7 , c = 18$
$y = 2 x - 7$